Walled Projection

Let's call $v_x$ and $v_y(t)$ the two orthogonal projections of velocity $\vec {v(t)}$ .

Remember that $v_y$ depends on $t$ (so $v_y = v_y(t)$ ) while $v_x$ is constant.

Note that $v_x = \dfrac{50}{2,5} = 20 \dfrac{\textrm{m}}{\textrm{s}}$

Call $t_1$ the time when the object passes on top of the first wall. If the object was projected with velocity $\vec{v(t_1)}$ from the top of the first wall, it would then land on the top of the second wall so that its range would be $50$ m.

Using this fact and the formula for the range, we can know the value of $v_y(t_1)$ , i.e. the vertical velocity when the object reaches the top of the first wall.

In fact we have

$r = 2v_x v_y(t_1) / g$

$v_y(t_1) = \dfrac{g r}{2v_x} = \dfrac{10 \cdot 50}{2 \cdot 20} = \dfrac{25}{2}\dfrac{\textrm{m}}{\textrm{s}}$

By conservation of energy now we can find the vertical velocity $v_y(0)$ when the projectile is projected from the ground.

$\displaystyle{\dfrac{1}{2}m \left( v_y(0)\right) ^2 = \dfrac{1}{2}m \left( v_y(t_1)\right) ^2 + mgh}$

where $h = 7,5$ m.

Eliminating $m$ and solving we get

$v_y(0) = \dfrac{35}{2} \dfrac{\textrm{m}}{\textrm{s}}$

Using again the range formula

$\displaystyle{r = \dfrac{2v_x v_y(0)} {g} = \dfrac{2\cdot 20 \cdot \dfrac{35}{2}}{10} = \boxed{70 \textrm{m}} }$